Solveeit Logo

Question

Question: If for a spherical mirror object distance, \(u = (50.1 \pm 0.5)cm\) and image distance\(v = (20.1 \p...

If for a spherical mirror object distance, u=(50.1±0.5)cmu = (50.1 \pm 0.5)cm and image distancev=(20.1±0.2)cmv = (20.1 \pm 0.2)cm, then the focal length of spherical mirror will be
A. (14.3±0.1)cm(14.3 \pm 0.1)cm
B. (14.3±0.5)cm(14.3 \pm 0.5)cm
C. (30.1±0.1)cm(30.1 \pm 0.1)cm
D. (25.3±0.5)cm(25.3 \pm 0.5)cm

Explanation

Solution

Hint Find the focal length using mirror formula using object distance and image distance. Main part is error calculation, for this must know the error calculation in addition, multiplication and division. Use the formula 1f=1u+1v\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v} to find the value of focal length and to find value of error Δf\Delta fmodify the above formula and use that. Final answer is focal length with error.

Complete step-by-step solution :
Given u=(50.1±0.5)cmu = (50.1 \pm 0.5)cm and v=(20.1±0.2)cmv = (20.1 \pm 0.2)cm.
Putting these values in formula 1f=1u+1v\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v} or f=u.vu+vf = \dfrac{{u.v}}{{u + v}} we get,
f=(50.1)×(20.1)(50.1)+(20.1)=1007.0170.2=14.3cmf = \dfrac{{(50.1) \times (20.1)}}{{(50.1) + (20.1)}} = \dfrac{{1007.01}}{{70.2}} = 14.3cm
From 1f=1u+1v\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v},
Δff2=Δuu2+Δvv2\dfrac{{\Delta f}}{{{f^2}}} = \dfrac{{\Delta u}}{{{u^2}}} + \dfrac{{\Delta v}}{{{v^2}}} or Δf=Δu(f2u2)+Δv(f2v2)\Delta f = \Delta u\left( {\dfrac{{{f^2}}}{{{u^2}}}} \right) + \Delta v\left( {\dfrac{{{f^2}}}{{{v^2}}}} \right)
Δf=0.5(14.350.1)2+0.2(14.320.1)2\Delta f = 0.5{\left( {\dfrac{{14.3}}{{50.1}}} \right)^2} + 0.2{\left( {\dfrac{{14.3}}{{20.1}}} \right)^2}
Δf=0.04+0.1=0.14cm\Delta f = 0.04 + 0.1 = 0.14cm
Then f=14.3±0.1f = 14.3 \pm 0.1.

Hence the correct answer is option A.

Note:- Error for a±Δab±Δb\dfrac{{a \pm \Delta a}}{{b \pm \Delta b}} id given by (Δaa+Δbb)×(ab)\left( {\dfrac{{\Delta a}}{a} + \dfrac{{\Delta b}}{b}} \right) \times \left( {\dfrac{a}{b}} \right) ,error for (a±Δa)+(b±Δb)(a \pm \Delta a) + (b \pm \Delta b) is given by ±(Δa+Δb) \pm (\Delta a + \Delta b) and for (a±Δa)×(b±Δb)(a \pm \Delta a) \times (b \pm \Delta b) is given by ±(Δa×b+Δb×a) \pm (\Delta a \times b + \Delta b \times a). Focal length is positive then the given mirror is concave mirror because concave mirror has positive focal length and focal length of convex mirror is negative